What are the three ways to solve a quadratic equation?

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- Factorization
- Completing the square
- The quadratic formula

What are the steps to solving a quadratic equation by factorising?

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- Get everything on one side, leaving just 0 on the other. (i.e. x
^{2}-2x+1=0).
- Factorise, so you get something like (x+a)(x+b)=0.
- Solve each parenthesis for x. (i.e. x+a=0 and x+b=0)

^{2}-2x+1=0).

How can you recognize that an equation is quadratic?

Some variable is squared.

It can always be written as:

ax^{2}+bx+c=0

What are the steps for factorising a quadratic equation?

ax^{2}+bx+c=0

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- Factor out any common factors. (So anything that goes into all of the terms. i.e. 6x
^{2}-2x-8 = 2 (3x^{2}-x-4))
- Make a list of all the factors that multiply to make
*ac*.
- Check to see which two factors from step two add to the
*b* term.
- Rewrite the equation, separating the middle term into the two factors you found.
- Take the common factors out of the first two terms and the last two, and regroup.

^{2}-2x-8 = 2 (3x^{2}-x-4))*ac*.*b*term.For example, 3x^{2}-x-4:

- 6x
^{2}-2x+8=2(3x^{2}-x-4). Now I need to factorise 3x^{2}-x-4 - 3•-4=-12, so the factors are 1•-12, -1•12, 2•-6, -2•6, 3•-4, -3•4
- Do any of the pairs add to -1? Yep, 3+-4=-1.
- 3x
^{2}+3x-4x-4 - 3x(x+1) - 4(x+1) = (3x-4)(x+1). So the answer is 6x
^{2}-2x+8=2(3x-4)(x+1).

What does the graph of f(x)=x^{2} look like? What are the vertex and line of symmetry?

Vertex: (0,0)

Line of symmetry: x=0

How do you shift f(x)=x^{2} to the right by h?

You put a "-h" in with the x term.

g(x)=(x-h)^{2}

How do you shift f(x)=x^{2} to the up by k?

You put a "+k" at the end.

g(x)=x^{2}+k

How do you mirror f(x)=x^{2}+k over the x-axis?

You multiply the whole function by -1.

g(x)= -(x^{2}+k).

How do you mirror f(x)=x^{2}+k over the y-axis?

You multiply the x-part by -1.

g(x)=(-x)^{2}+k.

How do you vertically stretch f(x)=x^{2}+k by a?

You multiply the whole function by a.

g(x)=a(x^{2}+k).

How do you horizontally stretch f(x)=(x-h)^{2}+k by b?

You multiply the whole x-part by a.

g(x)=(b(x-h))^{2}+k.

Where is the vertex and line of symmetry for

f(x)=a(b(x-h)^{2}+k)?

Vertex: (h, a•k)

Line of symmetry: x=h

How can you find the root if a quadratic equation has "two equal real roots"?

This means the discriminant is zero, so your solution is just